I am trying to prove that if $r_i \sim Lap(0,1/\varepsilon)$ where $\varepsilon >0$ then:
$$Pr[r_i \geq 1+r^*] \geq e^{-\varepsilon}Pr[r_i \geq r^{*}]$$.
I know that for $r*>0$ it satisfies with equality. Even though, for $r <0$, I couldn't find out how to prove it.
Note $Lap \sim (\mu,b)$:
$$Pr[X \geq x] = 1-F(x)=\begin{cases} 1-\frac{1}{2}\exp(\frac{x-\mu}{b}) && \text{if }x< \mu \\ \frac{1}{2}\exp(-\frac{x-\mu}{b}) &&\text{if } x\geq \mu\end{cases},$$
OK, I did this quickly. Hope it’s correct.
When $r*\geq 0,$ the relationship holds as you observed. And when $r^*\leq -1,$ the same expression for both probabilities you want to compare enables a direct proof.
Let $r^*\in(-1,0),$ so that $1+r^* \in (0,1).$ Then what you want to show is $$\frac{1}{2} e^{-\epsilon(1+r^*)}\geq e^{-\epsilon}\left(1-\frac{1}{2} e^{\epsilon r^*}\right) $$ or $$ \frac{1}{2} e^{-\epsilon(r^*+1)}+ \frac{1}{2} e^{\epsilon (r^*-1)} \geq e^{-\epsilon} $$ or $$ e^{-\epsilon} \left( \frac{ e^{-\epsilon r^*}+ e^{\epsilon r^*}}{2} \right)\geq e^{-\epsilon} $$ which holds since the cosh function is lower bounded by $1.$
When $r*<-1$, a litlle more larger, we want to find the next inequality:
\begin{equation*} \begin{split} e^{\epsilon} \left(1-\frac{1}{2}e^{\epsilon(x+1)} \right) \geq 1-\frac{1}{2}e^{\epsilon(x)}\\ e^{\epsilon}-\frac{1}{2}e^{\epsilon x+ 2\epsilon}\geq 1-\frac{1}{2}e^{\epsilon(x)} \end{split} \end{equation*}
Which we will bounded by both inequalities using the fact that $r*\leq-1$
\begin{equation*} \begin{split} r* &\leq -1\\ e^{\epsilon r*} &\leq e^{-\epsilon}\\ e^{\epsilon x + 2\epsilon} &\leq e^{\epsilon}\\ -\frac{1}{2}e^{\epsilon r*+ 2\epsilon} &\geq -\frac{1}{2}e^{\epsilon}\\ e^{\epsilon}-\frac{1}{2}e^{\epsilon r*+ 2\epsilon} &\geq e^{\epsilon}-\frac{1}{2}e^{\epsilon}\\ \end{split} \end{equation*}
The same way we have:
\begin{equation*} \begin{split} r* &\leq -1\\ 1-\frac{1}{2}e^{\epsilon x} & \geq 1-\frac{1}{2}e^{-\epsilon} \end{split} \end{equation*}
Joining this inequalitys, we can obtain
\begin{equation*} \begin{split} e^{\epsilon}-\frac{1}{2}e^{\epsilon} > 1-\frac{1}{2}e^{-\epsilon}\\ 2(e^{\epsilon}-1) > e^{\epsilon}-e^{-\epsilon} \end{split} \end{equation*}
Where the inequalitie holds since $2>1$ and $-1 \leq - e^{-\epsilon}$
